2z^2=7z+5

Solution for 2z^2=7z+5 equation:

2z^2=7z+5

We move all terms to the left:

2z^2-(7z+5)=0

We get rid of parentheses

2z^2-7z-5=0

a = 2; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·2·(-5)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{89}}{2*2}=\frac{7-\sqrt{89}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{89}}{2*2}=\frac{7+\sqrt{89}}{4}
$